Find $\lambda^{d}(A_{d}(a))$ and determine its behaviour as $d \to \infty$

Question

Set $A_{d}(a):=\{(x_{1},...,x_{d})\in\mathbb [0,\infty[^{d}: \sum_{i=1}^{d}x_{i}\leq a\}$, whereby $a > 0$

Determine:

(i) $\lambda^{d}(A_{d}(a))$

(ii) how $(A_{d}(1))$ behaves as $d \to \infty$

Is there a formula for n-dimensional squares? (At least I think it is a square).

Answer 1

You can evaluate the volume as an iterated integral.

Since $x_1 + \cdots + x_d \le a$ a point $x = (x_1,\ldots,x_d)$ belongs to the region $A_d(a)$ if and only if

• $ 0 \le x_1 \le a$
• $ 0 \le x_2 \le a - x_1$
• $0 \le x_3 \le a - x_1 - x_2$

and so on until you get to

• $0 \le x_d \le a - x_1 - \cdots - x_{d-1}$.

The elementary $d$-volume of $A_d(a)$ satisfies $$V_d(A_d(a)) = \int_0^a \int_0^{a - x_1} \int_0^{a - x_1 - x_2} \cdots \int_0^{a - x_1 - \cdots - x_{d-1}} \, dx_d dx_{d-1} \cdots dx_1.$$

You could try to work this out directly, or look for a pattern and formulate a proof by induction. It isn't hard to see that

• $V_1(A_1(a)) = a$
• $V_2(A_2(a)) = \dfrac{a^2}{2}$
• $V_3(A_3(a)) = \dfrac{a^3}{6}$

so one might have the audacity to conjecture that $V_d(A_d(a)) = \dfrac{a^d}{d!}$.

Since the cases $d=1$ through $d=3$ have been dispensed with it remains only to make the induction step. Assuming that the formula holds for $d-1$, a close look at the integral defining $V_d(A_d(a))$ shows $$V_d(A_d(a)) = \int_0^a V_{d-1}(A_{d-1}(a - x_1)) \, dx_1 = \int_0^a \frac{(a-x_1)^{d-1}}{(d-1)!} \, dx_1 = \frac{a^d}{d!}.$$